How do I calculate the volume $V$ of the subset $G$ of $\Bbb R^3$, which is obtained when cylinder $x^2+y^2 \leq 1$ is cut from sphere $x^2+y^2+z^2 < 4$, i.e.,
$$G := \begin{Bmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix} & \in \mathbb{R}^3: x^2+y^2+z^2<4, \,\, x^2+y^2>1 & \end{Bmatrix}$$
I need to calculate
$$V=\displaystyle\int_{G} 1 \, \mathrm d x \, \mathrm d y \, \mathrm dz$$
How?
Maybe I need to use cylindrical coordinates by the transformation and integration...
$$\int_{x=-1}^{1}\int_{x=-\sqrt(1-x^2)}^{x=\sqrt(1-x^2)}\int_{x=-\sqrt(f-x^2-y^2)}^{x=\sqrt(4-x^2-y^2)} \, \mathrm d x \, \mathrm d y \, \mathrm dz$$
and than using transformation $x=rcos\theta, y=rsin\theta,z=z$ i get that volume is $4\pi\sqrt{3}$
Yes, you are on the right track. By using cylindrical coordinates, the volume is given by $$V=\int_{\theta=0}^{2\pi}\left(\int_{\rho=1}^2 \left(\int_{z=-\sqrt{4-\rho^2}}^{\sqrt{4-\rho^2}}1dz\right)\rho d\rho\right) d\theta =2\int_{\theta=0}^{2\pi}\left(\int_{\rho=1}^2 \sqrt{4-\rho^2}\rho d\rho\right) d\theta\\=4\pi\int_{\rho=1}^2 \sqrt{4-\rho^2}\rho d\rho = \frac{4\pi}{3}\left[-(4-\rho^2)^{3/2}\right]_1^2=4\pi\sqrt{3}.$$